To solve this problem, let's analyze the given equations [tex]\(x^2 + a^2 = 1 - 2ax\)[/tex] and [tex]\(x^2 + b^2 = 1 - 2bx\)[/tex].
We start by recognizing that both equations share a common root, let's denote this common root by [tex]\(x\)[/tex]. Because [tex]\(x\)[/tex] is a root of both equations, we can set the left-hand sides of both equations equal to each other and similarly the right-hand sides as well.
First Equation:
[tex]\[x^2 + a^2 = 1 - 2ax\][/tex]
Second Equation:
[tex]\[x^2 + b^2 = 1 - 2bx\][/tex]
Since both [tex]\(x^2\)[/tex] terms are identical, we equate the remaining terms on each side of the equations:
[tex]\[a^2 + 2ax = b^2 + 2bx\][/tex]
This simplifies to:
[tex]\[a^2 - b^2 = 2bx - 2ax\][/tex]
We can factor both sides of the equation:
[tex]\[(a^2 - b^2) = (2x)(b - a)\][/tex]
Note that [tex]\(a^2 - b^2\)[/tex] can be factored as a difference of squares:
[tex]\[(a - b)(a + b) = 2x(b - a)\][/tex]
As long as [tex]\(a \neq b\)[/tex], we can divide both sides by [tex]\((a - b)\)[/tex], simplifying to:
[tex]\[(a + b) = -2x\][/tex]
Now, recalling that the common solution [tex]\(x\)[/tex] applies to both equations, we identify the situation where [tex]\((a - b) = \pm 1\)[/tex]. Given the modulus condition, we find:
[tex]\[ |a - b| = 1 \][/tex]
Therefore, the correct option among the list provided is:
[tex]\[ \boxed{|a - b| = 1} \][/tex]
This is option (b).
